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2 Inversion Formula

In this section, we review the inversion formulae in the probability theory and the Fourier analysis. In the probability theory, the following L´evy’s inversion formula is most famous. Theorem 1 (L´evy’s inversion formula). Let X be a random variable and ϕ(t) be the characteristic function of X. Then, for a

R ita itb 1 e− e− 1 1 lim − ϕ(t)dt = P (X = a)+ P (a

1 (0

2 If f is a slowly increasing function, Equation (4) can be justified as an equation of Schwartz distributions. The following proposition is a justification of the intuitive discussion in Section 1 by a summation method.

Proposition 1. Let f be a probability density function and ϕ(t) := ∞ eityf(y)dy be the characteristic function of f. For any continuous point x of f, the equation −∞ R 0 1 i(x+iε)t ∞ i(x iε)t f(x)= lim e− ϕ(t)dt + e− − ϕ(t)dt 2π ε +0 0 → Z−∞ Z  holds.

We utilize a fact in [5, p28, Note 1.3] with a small change. Lemma 1. Let K(x) and ϕ(x) be real-valued functions on R. Suppose that K(x) is absolutely integrable and ∞ K(x)dx =1 holds, and that ϕ(x) is continuous ans bounded. For a positive number ε> 0, put −∞ R 1 t x ϕε(x) := K − ϕ(t)dt. ε R ε Z  

Then, for any x R, we have ϕ(x) = limε +0 ϕε(x). ∈ → Proof. Since ϕε(x) is decomposed as

∞ x t dt ∞ x t dt ϕ (x)= ϕ(x) K − (ϕ(x) ϕ(t)) K − ε ε ε − − ε ε Z−∞   Z−∞   ∞ x t dt = ϕ(x) (ϕ(x) ϕ(t)) K − , − − ε ε Z−∞   it is enough to show that the second term converges to zero. We decompose the integral domain into x t ε and x t <ε. On the first domain, we have | − |≥ | − | x t dt x t dt (ϕ(x) ϕ(t)) K − 2 sup ϕ(x) K − x t ε − ε ε ≤ x | | x t ε ε ε Z| − |≥     Z| − |≥  

= 2 sup ϕ(x) K (s) ds. x | | s 1/√ε | |   Z| |≥ Note that sup ϕ(x) is finite since ϕ(x) is bounded. Since K(x) is absolutely integrable, K (s) ds x | | s 1/√ε | | converges to 0 as ε +0. Consequenty, we have | |≥ → R x t dt (ϕ(x) ϕ(t)) K − 0 (ε +0). x t ε − ε ε → → Z| − |≥  

On the second domain, we have

x t dt (ϕ(x) ϕ(t)) K − sup ϕ(x) ϕ(t) K (s) ds x t <ε − ε ε ≤ x t <ε | − |! s 1/√ε | | Z| − |   | − | Z| |≥

∞ sup ϕ(x) ϕ(t) K (s) ds. ≤ x t <ε | − | | | | − | ! Z−∞

Since ϕ(x) is continuous, sup x t <ε ϕ(x) ϕ(t) goes to zero as ε +0. Consequently, we have | − | | − | → x t dt (ϕ(x) ϕ(t)) K − 0 (ε +0). x t <ε − ε ε → → Z| − |  

Therefore, we have ϕ(x) = limε +0 ϕε(x). →

3 Proof of Proposition 1. Fix the continuous point x of f. Decompose f into a sum of two continuous functions f1 and f2 where f1 is support compact and f2 equals to zero on a neighborhood of x. Put ity ϕj (t) := ∞ e fj(y)dy (j =1, 2). By the Fubini’s theorem, we have −∞ R 0 1 i(x+iε)t ∞ i(x iε)t e− ϕ1(t)dt + e− − ϕ1(t)dt 2π 0 Z−∞ Z  0 1 ∞ i(y x iε)t ∞ ∞ i(y x+iε)t = f1(y) e − − dtdy + f1(y) e − dtdy 2π 0 Z−∞ Z−∞ Z−∞ Z  ∞ 1 ε = f (y) dy 1 π (y x)2 + ε2 Z−∞ − 1 Apply Lemma 1 with K(x)= π(x2+1) , then we have

∞ 1 ε f (y) dy f (x) (ε +0). 1 π (y x)2 + ε2 → 1 → Z−∞ − Analogously, we have

0 1 i(x+iε)t ∞ i(x iε)t 1 ∞ 1 1 e− ϕ2(t)dt + e− − ϕ2(t)dt = f2(y) dy 2π 0 2π i(y x iε) − i(y x + iε) Z−∞ Z  Z−∞  − − −  Since f2 equals to zero on a neighborhood of x, for sufficiently small δ > 0, the above integral equals to 1 1 1 f2(y) dy 2π y x >δ i(y x iε) − i(y x + iε) Z| − |  − − −  By the Lebesgue’s dominated convergence theorem, this integral converges to 0 as ε +0. Hence, we have → 0 1 i(x+iε)t ∞ i(x iε)t lim e− ϕ(t)dt + e− − ϕ(t)dt 2π ε +0 0 → Z−∞ Z  2 0 1 i(x+iε)t ∞ i(x iε)t = lim e− ϕj (t)dt + e− − ϕj (t)dt = f1(x)= f(x). 2π ε +0 j=1 → 0 X Z−∞ Z 

3 Perspective of Hyperfunction

In this section, we briefly review the theory of hyperfunctions and justify the intuitive discussion in Section 1 from the view point of hyperfunctions. In the first, we review the theory of hyperfunctions of a single variable based on [11] and [5]. Let be the sheaf of holomorphic functions on the complex plane C. The sheaf of the hyperfunctions ofO a 1 B single variable is defined as the 1-th derived sheaf R( ) of . The global section (R) is the inductive limit lim (U R)/O(U) with respect to the familyH O of complexO neighborhoods UB R. The elements U R O \ ⊃ of (R−→) are⊃ called hyperfunctions on R. Any hyperfunction on R can be represented as a equivalent classB [F (z)] with a representative F (z) (U R). ∈ O \ Let L1,loc(R) be the spaces of locally integrable functions on R. There is a natural embedding ι from L1,loc(R) to (R). When an integrable function f L1(R) satisfies some condition, we can take a representative F (zB) of ι(f) as a holomorphic function on∈C R. \ Proposition 2 (M. Sato). Let f(x) be an integrable function on R such that the integral ∞ (x2 + 1 −∞ 1)− f(x)dx is finite. Take a constant α C R, and put ∈ − R 1 ∞ 1 1 F (z)= f(x) dx (z C R). 2πi x z − x α ∈ \ Z−∞  − −  Then we have ι(f) = [F ].

4 Proof. See, [11]. We call a measure µ on R to be locally integrable when µ(K) is finite for any compact set K R. ⊂ We denote by Mloc the space of the locally integrable measures on R. Analogous to the case of L1,loc(R), there is a natural embedding ι from M to (R), and the following proposition holds: loc B 2 1 Proposition 3 (M. Sato). Let µ be a measure on R such that the integral R(x + 1)− µ(dx) is finite. Take a constant α C R, and put ∈ − R 1 1 1 F (z)= µ(dx) (z C R) 2πi R x z − x α ∈ \ Z  − −  Then we have ι(µ) = [F ].

Let I be a neighborhood of 0 R, and put I∗ := I 0 . A function F (z) holomorphic on a tubular ∈ \{ } domain R + iI∗ := z C z I∗ is said to be slowly increasing if for any compact subset K I∗ and any ε> 0, there exists{ ∈C >|ℑ0 such∈ } that ⊂ ε z F (z) Ce |ℜ | | |≤ uniformly for z R + iK. A hyperfunction f on R is said to be slowly increasing if we can take a slowly increasing function∈ F (z) as its representative, i.e., there exists a slowly increasing function F (z) such that f = [F (z)]. Put

z z e e− χ+(z) := z , χ (z) := z . 1+ e − 1+ e− The following definition is a specialization into the case of a single variable of the general definition in [5].

Definition 1. Let a slowly increasing function F (z) on a tubular domain R + iI∗ be a representative of a slowly increasing hyperfunction f (R). Take a positive number y I∗. We define the Fourier transform of f = [F (z)] as f = [G(w)]∈ where B ∈ F iwz iwz e− χ (z)F (z)dz e− χ (z)F (z)dz ( w> 0) z=+y − z= y − G(w)= ℑ iwz − ℑ − iwz ℑ (R z=+y e− χ+(z)F (z)dz +R z= y e− χ+(z)F (z)dz ( w< 0). − ℑ ℑ − ℑ Here, for y R and a functionR ϕ(z) with a complex variableR z, we denote the integral ϕ(x + iy)dx by ∈ R z=y ϕ(z)dz. ℑ R R In the second, we give and show an equation corresponding to the formal equation (2). Let X be a itX random variable. The distribution µX of X and the characteristic function ϕX (t) = E e of X can be regarded as hyperfunctions. We denote them by ιµ and ιϕ respectively. The following equation x X
corresponds to the formal equation (2).

(ιϕ )=2πιµ . (6) F X X Since the characteristic function can be regarded as the inverse Fourier transformation of the distribution under a suitable condition, the equation (6) is an inversion formula. We calculate defining functions of ιµX and ιϕX . Let α C R be a constant. Since µX is a probability measure on R, we have ∈ \ 1 1 2 µX (dx) 2 µX (dx) µX (dx)=1 < R (x + 1) ≤ R x +1 ≤ R ∞ Z Z Z | | and 1 1 1 1 µX (dx) µX (dx) µX (dx)= < . R x α ≤ R x α ≤ R α α ∞ Z Z Z − | − | |ℑ | |ℑ | By Proposition 3, we have

1 1 1 1 1 ιµX = µX (dx) = µX (dx) . (7) 2πi R x z − x α 2πi R x z  Z  − −    Z − 

5 2 1 Since the estimation ∞ ϕX (t)(t + 1)− dt < holds, we can apply Proposition 2 to ϕX (t). In fact, −∞ ∞ we have R

∞ ϕX (t) ∞ ϕX (t) ∞ 1 ∞ 2 dt 2 dt 2 dt = [arctan(t)] = π. t +1 ≤ t +1 ≤ t +1 −∞ Z−∞ Z−∞ Z−∞

Hence, we have 1 ∞ 1 1 ιϕ = ϕ (t) dt . X 2πi X t w − t α  Z−∞  − −   We calculate the Fourier transformation of ιϕX .

Lemma 2. For a random variable X, the hyperfunction ιϕX corresponding to the characteristic function ϕX of X is slowly increasing. Proof. For simplicity, we assume α = √ 1 without loss of generality. The absolute value of the repre- − sentative of ιϕX satisfies the inequality

1 ∞ 1 1 w α ∞ dt ϕ (t) dt | − | . (8) 2πi X t w − t α ≤ 2π t w t α Z−∞   Z−∞ − − | − || − | Put c := w , then the inequalities |ℜ | (c+1) c 1 ∞ dt ∞ dt − dt − − dt 2 =1, 2 =1, c+1 t w t α ≤ c+1 (t c) t w t α ≤ (t + c) Z | − || − | Z − Z−∞ | − || − | Z−∞ c+1 dt 2(c + 1) , c 1 t w t α ≤ w Z− − | − || − | |ℑ | imply that the right hand side of (8) is bounded above by

w α w +1 | − | 1+ |ℜ | . π w  |ℑ | 

Hence, ιϕX is a slowly increasing hyperfunction.

In order to calculate an explicit form of a defining function of the Fourier transformation of ιϕX , we decompose the function ϕX (t) as

ϕ (t)=(1 H(t))ϕ (t)+ H(t)ϕ (t) X − X X where H(t) is the Heaviside function. Embedding the both sides of the above equation into the global section B(R) of the sheaf of hyperfunctions, we have

ιϕ = ι((1 H)ϕ )+ ι(Hϕ ) X − X X By Proposition 2, defining functions of ι((1 H)ϕ ) and ι(Hϕ ) are given by − X X 1 0 1 1 ϕ1(w) := ϕX (t) dt (w C R 0), 2πi t w − t α ∈ − ≤ Z−∞  − −  1 ∞ 1 1 ϕ2(w) := ϕX (t) dt (w C R 0) 2πi t w − t α ∈ − ≥ Z0  − −  respectively. Then we have ιϕX = [ϕ1(w)] + [ϕ2(w)]. Analogous to Lemma 2, we can show that both [ϕ1(w)] and [ϕ2(w)] are slowly increasing hyperfunctions. (0+) For a function F (z) with a complex variable , we denote by + F (z)dz the integral along the path (0+) ∞ described in Figure 1. We also denote by F (z)dz the integralR along the path described in Figure 2. We denote by H the upper half-plane −∞z C z > 0 . R{ ∈ |ℑ }

6 O O

(0+) Figure 1: Contour of the path of + (0+) ∞ Figure 2: Contour of the path of R −∞ R Lemma 3. The holomorphic functions

(0+) izw e− ϕ1(w)dw (z H), ψ1(z) := − −∞ ∈ (0 (z H), R ∈− 0 (z H), ψ (z) := ∈ 2 (0+) izw ( + e− ϕ2(w)dw (z H), ∞ ∈− R are defining functions of the Fourier transformations of [ϕ1(w)] and [ϕ2(w)] respectively, i.e., we have [ϕ (w)] = [ψ (z)] (k =1, 2). F k k Proof. By Definition 1, a defining function of [ϕ (w)] is given by F 1 izw izw e− χ (w)ϕ1(w)dw e− χ (w)ϕ1(w)dw ( z > 0) ˜ w=+y − w= y − ψ1(z)= ℑ izw − ℑ − izw ℑ (R w=+y e− χ+(w)ϕ1(w)dw +R w= y e− χ+(w)ϕ1(w)dw ( z < 0). − ℑ ℑ − ℑ R R Since ϕ1 is holomorphic on C R 0, we have − ≤ (0+) izw e− χ (w)ϕ1(w)dw ( z > 0) ψ˜ (z)= − − ℑ 1 (0+)−∞ izw ( R e− χ+(w)ϕ1(w)dw ( z < 0). −∞ ℑ

(0+) izw R Note that the integral e− χ+(w)ϕ1(w)dw defines a holomorphic function on a neighborhood of R. Hence, −∞ R (0+) izw ψ (z)= ψ˜ (z) e− χ (w)ϕ (w)dw 1 1 − + 1 Z−∞ is a defining function of [ϕ1(w)] also. We can show [ϕ (wF)] = [ψ (z)] analogously. F 2 2 Lemma 4. Put 0 izt e− ϕX (t)dt (z H), ψ(z)= −∞ izt ∈ ∞ e− ϕ (t)dt (z H). (R− 0 X ∈− Then ψ(z) is a defining function of ιϕ . R F X

7 Proof. It is enough to calculate the right hand side of the equation ιϕX = [ψ1 + ψ2]. For z H, we have F ∈− (0+) izw 1 ∞ 1 1 ψ2(z)= e− ϕX (t) dtdw + 2πi 0 t w − t α Z ∞ Z  − −  (0+) 1 ∞ izw 1 1 = e− ϕX (t) dtdw 2πi + 0 t w − t α Z ∞ Z  − −  By the assumption z H, we have ∈− (0+) ∞ izw 1 1 e− ϕX (t) dtdw < . + 0 t w − t α ∞ Z ∞ Z   − − By the Fubini-Tonelli theorem and Cauchy’s integral formula, ψ (z ) equals to 2 (0+) 1 ∞ izw 1 1 e− ϕX (t) dwdt 2πi 0 + t w − t α Z Z ∞  − −  (0+) 1 ∞ izw 1 1 ∞ izt = ϕX (t) e− + dwdt = e− ϕX (t)dt. −2πi 0 + w t t α − 0 Z Z ∞  − −  Z 0 izt For z H, we can show ψ1(z)= e− ϕX (t)dt analogously. ∈ −∞ Lemma 5. For z C R, the followingR equation holds: ∈ \ 1 1 ψ(z)= µX (dx). (9) i R x z Z − Proof. For z H, we have ∈− ∞ izt ∞ i(z x)t e− ϕX (t)dt = e− − µX (dx)dt (10) − − R Z0 Z0 Z by the definition of the characteristic function. Since z is negative, we have ℑ ∞ i(z x)t ∞ zt 1 e− − µX (dx)dt = eℑ dt < . 0 R 0 ≤− z ∞ Z Z Z ℑ

By the Fubini-Tonelli theorem, the right hand side of (10) equals to ∞ i(z x)t 1 1 1 e− − dtµX (dx)= µX (dx)= µX (dx). − R R i(z x) i R x z Z Z0 Z − − Z − For z H, we can show the equation (9) analogously. ∈ Theorem 2. For a random variable X, equation (6) holds. Proof. By Lemma 5 and equation (7), we have 1 1 ιϕX = [ψ(z)] = µX (dx) =2πιµX . F i R x z  Z − 

Other Proof of Proposition 1. Let X be a random variable whose probability density function is f(x). Then, ϕ(t) equals to the characteristic function of X. By Lemma 2 and Lemma 4, the hyperfunction ιϕ corresponding to ϕ(t) is a Fourier hyperfunction, and a defining function of ιϕ is F 0 izt e− ϕ(t)dt (z H), ψ(z) := −∞ izt ∈ ∞ e− ϕ(t)dt (z H). (−R 0 ∈− 1 By Theorem 2, the corresponding hyperfunctionR ιf of f equals to (2π)− ιϕ. For any continuous point x of f, the boundary-value representation [5, Theorem 1.3.12] implies F 0 1 i(x+iy)t ∞ i(x iy)t f(x)= lim e− ϕ(t)dt + e− − ϕ(t)dt 2π y +0 0 → Z−∞ Z 

8 4 Sum of r-th Power of the Normal Variables

Fix a positive integer r 3. Let X1,...,Xn be the independent, identically and normally distributed ≥ n r r random variables, and put X := k=1 Xk . Let ϕ(t) be the characteristic function of X1 , i.e., put

P r r 1 2 itX1 itx 1 x ϕ(t)= E e = e e− 2 dx. (11) R √2π   Z n Then, the characteristic function ϕX of the powered sum X equals to ϕ(t) . By Proposition 1, we have

0 1 i(x+iy)t n ∞ i(x iy)t n fX (x)= lim e− ϕ(t) dt + e− − ϕ(t) dt (12) 2π y +0 0 → Z−∞ Z  for any continuous point x R of fX . In order to calculate the∈ limit in the right hand side of (12), we discuss on the analytic continuations of the each term of (3). In the first, we consider the characteristic function in (11). The characteristic function can be decom- posed as 0 1 ∞ 1 1 1 ϕ(t)= exp itxr x2 dx + exp itxr x2 dx. √2π 0 − 2 √2π − 2 Z   Z−∞   We are interested in the analytic continuation of each term in the right hand side. In order to calculate them, we consider the following general form of integral:

1 r 1 2 ϕθ(w) := exp iwz z dz (w C), (13) √2π − 2 ∈ Zγθ   where the integral path γ with parameter θ R is defined by θ ∈ γ (t) := teiθ (0

Note that we have ϕ(t)= ϕ0(t) ϕπ(t) for t R. For z C, put − ∈ ∈ zH := zw w H . { | ∈ } Examining the convergence region of the integral (13) for given parameter θ R, we obtain the following lemma: ∈

iθr Lemma 6. Suppose that d 3. For given parameter θ R, the integral (13) converges on e− H. ≥ ∈ iθr Proof. Take any point w e− H. By the straight forward calculation, we have ∈

1 ∞ iθr r 1 2iθ 2 iθ ϕθ(w) exp iwe t e t e dt | |≤ √2π − 2 Z0  

1 ∞ iθr r cos(2θ) 2 = exp (we )t t dt √2π −ℑ − 2 Z0   Since assumption weiθr H implies (weiθr) > 0, there exist t > 1 and ε> 0 such that ∈ ℑ 0

iθr cos(2θ) (we ) r 2 < ε −ℑ − 2t − − holds for any t>t0. Hence, we have

∞ cos(2θ) exp (weiθr)tr t2 dt −ℑ − 2 Zt0   ∞ r ∞ r 1 r exp( εt0) exp( εt ) dt r t − exp( εt ) dt = − . ≤ − ≤ · − ε Zt0 Zt0 iθr Consequently, the integral (13) converges on e− H.

9 iθr Lemma 7. Suppose that d 3. If cos2θ > 0 holds, then the integral (13) converges on e− H¯ . Here, H¯ denotes the closure of H.≥

iθr iθr Proof. For any point w e− H¯ , we have (we ) 0, This implies ∈ ℑ ≥

1 ∞ iθr r cos(2θ) 2 ϕθ(w) exp (we )t t dt | |≤ √2π −ℑ − 2 Z0   1 ∞ cos(2θ) exp t2 dt < ≤ √2π − 2 ∞ Z0  

Example 1. Let r = 3. Figures 3 and 4 show the contour of the integral path γθ and the region exp( iθr) H for θ = 0 respectively. Figures 5 and 6 show them for θ = π/9. −

γ

Figure 3: The integral path for θ =0 Figure 4: The convergence region for θ =0

′ iθr iθ r Lemma 8. If θ θ′ < π/d, then ϕ (w)= ϕ ′ (w) holds for w e− H e− H. | − | θ θ ∈ ∩ Proof. Put

′ ′′ iθ iθ iθ γ (t) := te (0 t R), γ (t) := te (0 t R), γ (θ′′) := Re (θ θ′′ θ′). 1,R ≤ ≤ 2,R ≤ ≤ 3,R ≤ ≤ By the Cauchy’s integral formula, we have

Φdz Φdz + Φdz =0. (15) − Zγ1,R Zγ2,R Zγ3,R 1/2 r 2 Here, we put Φ = (2π)− exp iwz z /2 . By some calculations, we have that there exist M > 0 and ε> 0 such that −
Φdz MR exp( εRr) . γ3,R ≤ − Z

Taking the limit of (15) as R approaches inifinity, we have ϕ (z) ϕ ′ (z)=0. θ − θ In the second, we disscuss on the analytic continuation of the each term of (3). Let us consider the following integral: iwz n ψ(z; θ,θ′, φ) := e− (ϕ (w) ϕ ′ (w)) dw. (16) θ − θ Zγφ where θ,θ′, φ R, z C, and the integral path γ is defined by (14). ∈ ∈ φ

10 γ θd

θ

Figure 5: The integral path for θ = π/9 Figure 6: The convergence region for θ = π/9

′ Lemma 9. Suppose (ei(φ+θr)) 0 and (ei(φ+θ r)) 0. Then the integral (16) converges for z exp( iφ)H. ℑ ≥ ℑ ≥ ∈ − − Proof. By the binomial expansion, the integral (16) equals to

n n n k iwz k n k ( 1) − e− ϕθ(w) ϕθ′ (w) − dw. k − γφ kX=0   Z With some calculation, the each term of the above expression can be written as

1 ∞ iwz k ′ n k e− ϕθ(w) ϕθ (w) − dw = n/2 exp( h1(z,t)s + h2(t)) dtds, (2π) Rn − Zγφ Z0 Z >0 where we put t := (t1,...,tn)⊤, dt := dt1 ...dtn,

k n k n ′ ′ iφ iθr r iθ r r 1 2iθ 2 2iθ 2 h1(z,t)= ie z e tℓ e tℓ , h2(t)= e tℓ + e tℓ . − − ! −2 ! Xℓ=1 ℓ=Xk+1 Xℓ=1 ℓ=Xk+1 By the Fubini-Tonelli theorem, it is enough to consider the convergence of the following integral:

∞ exp( h1(z,t)s + h2(t)) dsdt. (17) Rn | − | Z >0 Z0 For z exp( iφ)H, we have (izeiφ) > 0. This inequality and the assumption ( iei(φ+θr)) 0 and ′ ( ie∈−i(φ+θ r)) − 0 imply h (z,tℜ ) > 0. Hence, the integral (17) equals to ℜ − ≥ ℜ − ≥ ℜ 1 ∞ exp(h2(t)) exp(h2(t)) exp( s h1(z,t)) dsdt = | |dt. (18) Rn | | − ℜ Rn h1(z,t) Z >0 Z0 Z >0 ℜ By the assumptions, the integral (18) is bounded from above by

k n k 1 1 ∞ 1 2 ∞ 1 ′ 2 − 2 cos(2θ)t 2 cos(2θ )t iφ exp(h2(t)) dt = iφ e− dt e− dt (ize ) Rn | | (ize ) ℜ Z >0 ℜ Z0  Z0  k/2 (n k)/2 1 π π − = < . (izeiφ) 2 cos(θ) 2 cos(θ ) ∞ ℜ    ′  Therefore, the integral (16) converges.

11 ′ Lemma 10. Suppose (ei(φk+θr)) 0 and (ei(φk+θ r)) 0 for k =1, 2. The equation ℑ ≥ ℑ ≥

ψ(z; θ,θ′, φ1)= ψ(z; θ,θ′, φ2) holds for z ( exp( iφ )) ( exp( iφ )). ∈ − − 1 ∩ − − 2 Proof. Put

γ (t) := teiφ1 (0 t R), γ (t) := teiφ2 (0 t R), γ (φ) := Reiφ (φ φ φ ). 1,R ≤ ≤ 2,R ≤ ≤ 3,R 1 ≤ ≤ 2 By the Cauchy’s integral formula, we have

Φdw Φdw + Φdw =0, (19) − Zγ1,R Zγ2,R Zγ3,R iwz n where we put Φ = e− (ϕ (w) ϕ ′ (w)) . By some calculations, we have θ − θ

Φdw MR exp( R) γ3,R ≤ − Z

for sufficiently large M > 0. Taking the limit of (19) as R approaches inifinity, we have ψ(z; θ,θ′, φ ) 1 − ψ(z; θ,θ′, φ2)=0.

Theorem 3. Let ε> 0 be a sufficiently small positive number and x be a continuous point of fX . When r is odd, we have

ψ(x; ε, π, εr) ψ(x; ε,π,π + εr) (x> 0), fX (x)= − − − ψ(x;0, π ε,εr) ψ(x;0, π + ε, π εr) (x< 0). ( − − − When r is even, we have

ψ(x; ε, π + ε, εr) ψ(x; ε, π ε, π + εr) (x> 0), fX (x)= − − − − ψ(x;0,π,εr) ψ(x;0,π,π εr) (x< 0). ( − − Proof. When r is odd and x> 0, we have 1 fX (x)= lim (ψ(x iy;0, π, 0) ψ(x + iy;0,π,π)) y +0 2π → − − 1 = lim (ψ(x iy; ε, π, 0) ψ(x + iy;0, π + ε, π)) y +0 2π → − − 1 = lim (ψ(x iy; ε, π, εr) ψ(x + iy;0, π + ε, π + εr)) y +0 2π → − − − 1 = (ψ(x; ε, π, εr) ψ(x;0, π + ε, π + εr)) 2π − − When r is odd and x< 0, we have 1 fX (x)= lim (ψ(x iy;0, π, 0) ψ(x + iy;0,π,π)) 2π y +0 → − − 1 = lim (ψ(x iy; ε, π, 0) ψ(x + iy;0, π ε, π)) 2π y +0 → − − − − 1 = lim (ψ(x iy; ε,π,εr) ψ(x + iy;0, π ε, π εr)) y +0 2π → − − − − − 1 = (ψ(x; ε,π,εr) ψ(x;0, π ε, π εr)) 2π − − − −

12 When r is even and x> 0, we have 1 fX (x)= lim (ψ(x iy;0, π, 0) ψ(x + iy;0,π,π)) y +0 2π → − − 1 = lim (ψ(x iy; ε, π + ε, 0) ψ(x + iy;0, π + ε, π)) y +0 2π → − − 1 = lim (ψ(x iy; ε, π + ε, εr) ψ(x + iy;0, π + ε, π + εr)) 2π y +0 → − − − 1 = (ψ(x; ε, π + ε, εr) ψ(x;0, π + ε, π + εr)) 2π − − When r is odd and x< 0, we have 1 fX (x)= lim (ψ(x iy;0, π, 0) ψ(x + iy;0,π,π)) y +0 2π → − − 1 = lim (ψ(x iy; ε, π, 0) ψ(x + iy;0, π ε, π)) 2π y +0 → − − − − 1 = lim (ψ(x iy; ε,π,εr) ψ(x + iy;0, π ε, π εr)) 2π y +0 → − − − − − 1 = (ψ(x; ε,π,εr) ψ(x;0, π ε, π εr)) 2π − − − −

Remark 1. The all of integral representations in Theorem 3 make sense as Lebesgue integral. As we have shown in Lemma 9, when we write the integral in Theorem 3 into an integral on R, the integrand is a Lebesgue integrable function. Acknowledgements: We are grateful to K. Yajima, T. Oshima, and S.-J. Matsubara-Heo for valuable comments to Proposition 1. This work was supported by MEXT/JSPS KAKENHI Grand Numbers JP 25220001, 18J01507.

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